help please!!?

Question

If the graph of y = x^2 + ax + b has its vertex at (-1,4), how many real solutions are there for x^2 + ax + b = 3?

Answer 1

y(-1) = (-1)^2 + a(-1) + b = 4
-a + b = 4
y'(x) = 2x + a = 0 or at the vertex, x = -a/2 = -1, so a = 2
b = 4+a = 4+2 = 6
Real solns?
x^2 + 2x + 6 = 3
x^2 + 2x - 3 = 0
(x-1)(x+3) = 0
x = 1,-3 two real solns

Answer 2

Hey,

ok there are two ways of solving, one by graphical and second by arithmetic operations.

Graphical:
Draw the graph of x^2 + ax + b. Since the coefficient of x^2 is positive, its vertex at (-1,4) will be a minimum point.
Thus, the lowest value of y is 4.
Next, u draw the line y=3.
U will find that the graph of y = x^2 + ax + b and y = 3 doesnt coincide at all. That is, there is no solution for
x^2 + ax + b = 3.

Second method:
Solve for a and b first.
Lets solve for a first.
y = x^2 t ax + b------ (1)
=> dy/dx = 2x + a ------ (2)

At its vertex, dy/dx = 0
And since x = -1, we will have
From (2), 2(-1) + a = 0
=> a = 2
Now, substitute a = 2 into (1)
then, from (1), we will have y = x^2 + 2x + b------(3)
Since (-1, 4) is a point on the curve, substitute x=-1 and y=4
into (3)

from (3), we have 4 = (-1)^2 + 2(-1) + b
=> 4 = 1 - 2 + b
=> b = 5
Thus, the equation (1) will become
y = x^2 + 2x + 5
And now, we equate it to 3 i.e x^2 + 2x +5 = 3
=> x^2 + 2x +2 = 0
=> (x^2 + 2x + 1) + 1 = 0
=> (x+1)^2 + 1 = 0
=> (x+1)^2 = -1
which is not possible
Hence, there are no real solutions.

Answer 3

3

Answer 4

0 real solutions

Answer 5

Zero

Answer 6

There are two real solutions. There are no other solutions.

Good Luck!